If `asintheta+bcostheta=c` , then `acostheta-b sintheta` is equal to

Options

A. `+-sqrt(a+b+c)`
B. `+-sqrt(a^(2)+b^(2)+c^(2))`
C. `+-sqrt(c^(2)+a^(2)-b^(2))`
D. `+-sqrt(c^(2)+a^(2)-b^(2))`

Answer

To solve the problem where \( A \sin \theta + B \cos \theta = C \) and we need to find \( A \cos \theta - B \sin \theta \), we will follow these steps:

### Step 1: Square the first equation
We start with the equation:
\[ A \sin \theta + B \cos \theta = C \]
Squaring both sides gives:
\[ (A \sin \theta + B \cos \theta)^2 = C^2 \]
Expanding the left side, we have:
\[ A^2 \sin^2 \theta + 2AB \sin \theta \cos \theta + B^2 \cos^2 \theta = C^2 \]

### Step 2: Set up the second equation
Now, let's denote:
\[ X = A \cos \theta - B \sin \theta \]
We will square this expression as well:
\[ (A \cos \theta - B \sin \theta)^2 = X^2 \]
Expanding this gives:
\[ A^2 \cos^2 \theta - 2AB \sin \theta \cos \theta + B^2 \sin^2 \theta = X^2 \]

### Step 3: Add the two squared equations
Now we will add the two equations we have:
1. \( A^2 \sin^2 \theta + B^2 \cos^2 \theta + 2AB \sin \theta \cos \theta = C^2 \)
2. \( A^2 \cos^2 \theta + B^2 \sin^2 \theta - 2AB \sin \theta \cos \theta = X^2 \)

Adding these gives:
\[ (A^2 \sin^2 \theta + B^2 \cos^2 \theta + 2AB \sin \theta \cos \theta) + (A^2 \cos^2 \theta + B^2 \sin^2 \theta - 2AB \sin \theta \cos \theta) = C^2 + X^2 \]

### Step 4: Simplify the left side
Notice that:
- \( A^2 \sin^2 \theta + A^2 \cos^2 \theta = A^2 (\sin^2 \theta + \cos^2 \theta) = A^2 \)
- \( B^2 \sin^2 \theta + B^2 \cos^2 \theta = B^2 (\sin^2 \theta + \cos^2 \theta) = B^2 \)

Thus, we can simplify the left side:
\[ A^2 + B^2 = C^2 + X^2 \]

### Step 5: Rearranging to find \( X^2 \)
Now, rearranging gives:
\[ X^2 = A^2 + B^2 - C^2 \]

### Step 6: Solve for \( X \)
Taking the square root of both sides, we find:
\[ X = \pm \sqrt{A^2 + B^2 - C^2} \]

Thus, the final answer is:
\[ A \cos \theta - B \sin \theta = \pm \sqrt{A^2 + B^2 - C^2} \]